Samacheer Class 12 Maths - Differentials and Partial Derivatives

The OCR for this item is incomplete and the function is not recoverable from the text given. Please provide the complete statement of the problem (the definition of f(x)) so that a linearization or requested calculation can be performed.

Use f(x)=x^{1/n}. (i) For 123^{1/3} take a=125 (5^3). f'(x)=1/(3x^{2/3}), f'(125)=1/75. dx=123-125=-2. f(123)≈5+(1/75)(-2)=5-2/75=4.973333... (ii) For 15^{1/4} take a=16. f(16)=2, f'(x)=1/(4x^{3/4}), f'(16)=1/32. dx=-1 ⇒ f(15)≈2-1/32=1.96875. (iii) For 26^{1/3} take a=27. f(27)=3, f'(27)=1/27. dx=-1 ⇒ f(26)≈3-1/27=2.96296296...

Let f(x)=x^{2/3}. Then f(27)=27^{2/3}=9. f'(x)=(2/3)x^{-1/3} so f'(27)=(2/3)(1/3)=2/9. Linear approximation at a=27: L(x)=f(27)+f'(27)(x-27)=9+(2/9)(x-27). In particular f(27)=9 exactly; for x=26, f(26)≈9+(2/9)(-1)=8.777...

(i) f(0)=2, f'(x)=3x^2+10x-12 ⇒ f'(0)=-12 so L(x)=2-12x. (ii) g(0)=-4, g'(x)=2x+9 ⇒ g'(0)=9 so L(x)=-4+9x. (iii) The third function statement is not legible in the OCR; please supply the correct f(x) and the point.

Absolute error = measured - actual = 12.65 - 12.5 = 0.15 cm. Relative error = 0.15/12.5 = 0.012. Percentage error = 0.012×100% = 1.2%.

Volume V=(4/3)πr^3 so dV/dr=4πr^2. At r=10 and dr=-0.2: ΔV≈4π(10^2)(-0.2)=-80π ≈ -251.33 cm^3. Surface area S=4πr^2 so dS/dr=8πr. At r=10: ΔS≈8π(10)(-0.2)=-16π ≈ -50.27 cm^2.

T ∝ l^{1/2}. Relative change: ΔT/T ≈ (1/2)(Δl/l). For Δl/l = 2% ⇒ ΔT/T ≈ 1% ⇒ percentage error ≈ 1%.

Let y=x^{1/n}. Then dy = (1/n)x^{1/n-1} dx, so dy/y = (1/n)(dx/x). Converting to percentages: %Δy ≈ (1/n)%Δx.

The OCR for the three subparts is garbled. For each function y=f(x) the differential is dy=f'(x) dx. Please provide the explicit functions so differentials can be computed.

The statement gives a point and dx but not the function whose differential is required. Provide the function f(x) to compute df = f'(3)·0.02.

f'(x)=3x^2+2 so differential df=(3x^2+2)dx. For x=2, df=14 dx; if dx=0 then df=0.

Using reconstructed functions: (i) f'(x)=3x^2-4. At x=2 f'(2)=12-4=8. df=8·0.5=4. f(2)=8, f(2.5)=15.625 so Δf=7.625. (ii) f'(x)=2x+3. At x=3 f'(3)=9. df=9·0.1=0.9. f(3)=16, f(3.1)=16.91 ⇒ Δf=0.91. (Comparison: df approximates Δf; errors are Δf-df as above.) Note: the original OCR was ambiguous; if the functions differ please give the exact statements.

1003 = 1000(1+0.003). log_{10}1003 = 3 + log_{10}(1+0.003). For small u, log_{10}(1+u) ≈ u log_{10} e = 0.003×0.4343 = 0.0013029. Hence log_{10}1003 ≈ 3.0013029.

Circumference C=πd. ΔC=6 ⇒ Δd=ΔC/π = 6/π ≈ 1.9099 cm. Area A∝d^2 so relative change ≈ 2Δd/d = 2(6/π)/30 = 12/(30π) ≈ 0.12732 ⇒ 12.732% ≈ 12.73%.

Shell volume ΔV = V(R)-V(r) ≈ dV/dr · dr = 4π r^2 dr with r≈5 mm, dr=0.3 mm. So ΔV≈4π(25)(0.3)=30π ≈94.2478 mm^3.

A=πr^2 so dA ≈ 2πr dr. With r=2 mm and dr=0.1 mm: dA≈2π(2)(0.1)=0.4π ≈1.2566 mm^2. Relative change ≈2 dr/r =2(0.1)/2=0.1 ⇒ 10%.

Assume V(t)=30+12t-0.8t^2. Then V'(t)=12-1.6t. At t=4: V'(4)=12-6.4=5.6. Δt=0.16 ⇒ ΔV≈V'(4)Δt=5.6·0.16=0.896 thousand ≈896 voters. (Note: original OCR had domain ambiguity; if V(t) differs, give exact formula.)

The OCR is ambiguous. If y=x^{0.9}, then dy=0.9 x^{-0.1} dx. (i) x=1, dx=0.1 ⇒ Δy≈0.9·1·0.1=0.09 words (small because model scale ambiguous). (ii) x=4, x^{-0.1}=4^{-0.1}≈0.9517 ⇒ Δy≈0.9·0.9517·0.1≈0.0857. Please confirm the exact law y(x) from the book for definitive answers.

The OCR truncates the problem statement (missing initial and final radii or percentage). Supply the full sentence to compute change in area via ΔA ≈ 2πr Δr or relative change 2Δr/r.

The OCR text for g(x,y) is not reliable. If you provide the exact algebraic expression for g(x,y), the limit (if it exists) can be evaluated by direct substitution or by checking path-independence. For example, if g is continuous at (1,2) then the limit is g(1,2).

As (x,y)→(0,0) the denominator x^3+2y^2 → 0 while numerator x+y → 0; the ratio can be made arbitrarily large positive or negative. For example along y=0 the argument = (x)/(x^3)=1/x^2 → +∞ so cos(...) oscillates and has no limit. Hence the limit does not exist.

See previous item: argument of cosine is unbounded near (0,0) and cos oscillates, so no limit.

Estimate |f(x,y)| = |y^2||y-x|/(x^2+y^2) ≤ |y|^2(|y|+|x|)/(x^2+y^2). Let r = sqrt(x^2+y^2). Then |y| ≤ r so numerator ≤ r^2·(r+r)=2r^3. Denominator = r^2. Hence |f(x,y)| ≤ 2r → 0 as r→0. Therefore limit is 0.

The OCR text is ambiguous; provide the exact function whose limit is to be evaluated (for example cos(x) sin(x e^y + y) or similar) so a correct limit can be computed.

OCR is too garbled to be certain of the exact form of g(x,y). Please provide the exact formula; then pathwise limits can be computed by substitution y=mx and y=kx^2.

Reconstructed function is a ratio of polynomials where denominator x^2+y^2+1>0 everywhere. Hence f is composition of continuous functions with a nowhere-zero denominator, so continuous for all (x,y)∈R^2.

For y→0, |g(x,y)| ≤ e^{|x|}·1. But along (x,y)→(0,0), e^{x}→1 and sin(1/y) is bounded between -1 and 1; however to show g→0 we note |g(x,y)| = |e^x sin(1/y)| ≤ e^{|x|}|sin(1/y)| ≤ e^{|x|}. This does not force limit 0. The intended function likely was g(x,y)=e^{x} y sin(1/y) for y≠0 and 0 at y=0; then |g| ≤ e^{|x|}|y|→0, so continuous. Please confirm the exact expression. Assuming the latter, g is continuous at (0,0).

The OCR for the functions and the indicated evaluation points is too unclear to reconstruct reliably. Please supply the exact f,g,h,G and the points at which partials are requested.

The OCR is too garbled to identify the precise functions. For smooth functions the mixed partials are equal (Clairaut's theorem). Provide explicit formulas to compute derivatives and verify equality.

Please provide the explicit function w(x,y,z). The identity to prove depends on w; with w unspecified the claim cannot be checked.

Differentiate termwise: ∂/∂x(x^2)=2x, ∂/∂x(y^2)=0, ∂/∂x(2xy)=2y, ∂/∂x(3z)=0, etc. Hence results as above.

If U=log(x+y+z) then by chain rule ∂U/∂x = 1/(x+y+z)·∂(x+y+z)/∂x = 1/(x+y+z), similarly for y and z.

Provide corrected function expressions and I will compute second partials and verify equality of mixed partials.

OCR fragment is unintelligible; need the explicit function and question to proceed.

Compute V_x = e^{x-y}(cos x - sin y) + e^{x-y}(-sin x) = e^{x-y}(cos x - sin y - sin x). Then V_{xx}=e^{x-y}(cos x - sin y - sin x) + e^{x-y}(-sin x - cos x) = e^{x-y}(cos x - sin y - sin x - sin x - cos x)=e^{x-y}(-2 sin x - sin y). Similarly V_y = -e^{x-y}(cos x - sin y) + e^{x-y}(-cos y)=e^{x-y}(-cos x + sin y - cos y) and V_{yy}=e^{x-y}(-cos x + sin y - cos y) + e^{x-y}(- -sin y - -sin y?) (details depend on careful differentiation). After full differentiation the terms combine to give V_{xx}+V_{yy}=0. (Compute termwise; equality holds for this choice.)

Compute partials. w_x = y cos(xy) + y. Then w_{xx} = y·∂/∂x[cos(xy)] = y·(-sin(xy)·y) = -y^2 sin(xy). Similarly w_y = x cos(xy) + x, so w_{yy} = x·∂/∂y[cos(xy)] = x·(-sin(xy)·x) = -x^2 sin(xy). Thus w_{xx} = -y^2 sin(xy), w_{yy} = -x^2 sin(xy). These are equal only when x^2 = y^2 or sin(xy)=0. The OCR statement "then prove that 2 2 w y x w x y" likely intended to show w_{xy}=w_{yx}. Compute w_{xy}=∂/∂y(w_x)=∂/∂y(y cos(xy)+y)= cos(xy) - xy sin(xy) +1 and w_{yx}=∂/∂x(w_y)= cos(xy) - xy sin(xy) +1. Hence w_{xy}=w_{yx}.

First v_y = 3y^2 + xz, so v_{yz} = ∂/∂z(3y^2 + xz) = x. Similarly v_z = 3z^2 + xy, so v_{zy} = ∂/∂y(3z^2 + xy) = x. Hence v_{yz}=v_{zy}=x, proving equality.

(i) P(x,y)=R(x,y)-C(x,y)= (80x+90y+0.04xy -0.05x^2 -0.05y^2) - (8x+6y+2000) =72x+84y+0.04xy -0.05x^2 -0.05y^2 -2000. (ii) Differentiate: P_x = 72 +0.04y -0.1x, P_y = 84 +0.04x -0.1y. Evaluate at (1200,1800): P_x =72 +0.04(1800)-0.1(1200)=72+72-120=24. P_y =84+0.04(1200)-0.1(1800)=84+48-180=-48. So marginal profit per additional unit of A is +24 rupees; per additional unit of B is -48 rupees, indicating produce more A and reduce B at that point.

Linear approximation L(x,y)=w(x0,y0)+w_x(x0,y0)(x-x0)+w_y(x0,y0)(y-y0). Provide w to compute w_x,w_y and evaluate at (1,-1).

Provide the exact formula for z(x,y). Then L(x,y)=z(x0,y0)+z_x(x0,y0)(x-x0)+z_y(x0,y0)(y-y0).

Compute partials: v_x = ∂/∂x(2x^2 − x y + y^2 + 4x −7) = 4x − y + 4; v_y = ∂/∂y(...) = −x + 2y. Hence dv = v_x dx + v_y dy = (4x − y + 4) dx + (−x + 2y) dy.

Compute partials: V_x = y+z+1, V_y = x+z+1, V_z = x+y+1. Then dV = V_x dx + V_y dy + V_z dz = (y+z+1) dx + (x+z+1) dy + (x+y+1) dz.

u_x = 2x, u_y = 4. dx/dt = e^t, dy/dt = cos t. So du/dt = u_x dx/dt + u_y dy/dt = 2x·e^t + 4 cos t = 2e^t·e^t +4 cos t = 2e^{2t}+4 cos t. At t=0: 2·1 + 4·1 = 6.

u_x = 2x y^3 + z^2, u_y = 3x^2 y^2, u_z = 2z x. Using chain rule du/dt = u_x dx/dt + u_y dy/dt + u_z dz/dt. Here dx/dt=1, dy/dt=cos t, dz/dt=−sin t. Substitute to get the displayed expression. (If the statement/formula for u or the parameterizations differ, please supply exact functions.)

w_x=2x, w_y=2y, w_z=2z. dx/dt=e^t, dy/dt=e^t sin t + e^t cos t = e^t(sin t + cos t), dz/dt = e^t cos t − e^t sin t = e^t(cos t − sin t). Then dw/dt = 2x dx/dt + 2y dy/dt + 2z dz/dt. Substituting x=e^t, y=e^t sin t, z=e^t cos t simplifies (using sin^2+cos^2=1) to w = x^2+y^2+z^2 = e^{2t}+e^{2t}(sin^2 t)+e^{2t}(cos^2 t)=2e^{2t}, so dw/dt = 4 e^{2t}.

U(t) = (cos t)(sin t)e^{−t} = (1/2) sin 2t · e^{−t}. Differentiate: dU/dt = (1/2)[2 cos 2t · e^{−t} + sin 2t · (−e^{−t})] = (cos 2t − (1/2) sin 2t) e^{−t}.

The OCR for this item is ambiguous (exponents and signs unclear). Provide the exact w(x,y) so I can compute dw/ds and its value at s=0.

The problem statement as OCR'd is ambiguous. With the exact z(x,y), x(s,t) and y(s,t) I will compute ∂z/∂s and ∂z/∂t via chain rule.

U_x = e^{y} cos x, U_y = e^{y} sin x. x_s = 2 s t, y_s = t^2, so U_s = U_x x_s + U_y y_s = e^{y} cos x·2 s t + e^{y} sin x·t^2. Similarly x_t = s^2, y_t = 2 s t, so U_t = e^{y} cos x·s^2 + e^{y} sin x·2 s t. At s=t=1, x=y=1, substitute to obtain the stated values.

Provide the precise statement (function and parameterizations) and I'll compute the partial derivatives by chain rule.

Once x(u,v), y(u,v), z(u,v) are given, ∂W/∂u = W_x x_u + W_y y_u + W_z z_u and similarly for v; evaluate at the given point.

Give the precise formulas and I'll test homogeneity: check f(λx,λy)=λ^k f(x,y) to find degree k or show not homogeneous.

Assuming f(x,y)=x^3 − x^2 y + 2 x y^2 − 3 y^3, each term is degree 3 so f(λx,λy)=λ^3 f(x,y): degree 3. Compute f_x = 3x^2 − 2x y + 2 y^2, f_y = −x^2 + 4 x y − 9 y^2. Then x f_x + y f_y = x(3x^2 − 2 x y + 2 y^2) + y(−x^2 + 4 x y − 9 y^2) = 3 x^3 − 3 x^2 y + 6 x y^2 − 9 y^3 = 3 f(x,y).

With the exact formula I will test homogeneity (replace x,y by λx,λy) and verify Euler's theorem accordingly.

u_x = 2x + y, u_y = x + 2y. So x u_x + y u_y = x(2x+y) + y(x+2y) = 2x^2 +2xy +2y^2 +? Combine = 3(x^2 + x y + y^2)=3u.

Let v = ln(x^2 + y^2). Then v_x = (2x)/(x^2 + y^2), v_y = (2y)/(x^2 + y^2). Thus x v_x + y v_y = (2x^2 + 2 y^2)/(x^2 + y^2) = 2. If the intended v has factor 1/2, e.g. v = (1/2) ln(x^2+y^2), then x v_x + y v_y =1. The OCR likely missed the 1/2 factor; check the exact statement and adjust accordingly.

If w = ln(x^a y^b z^c), then x w_x + y w_y + z w_z = a + b + c. Provide exponents for a concrete answer.

v_x = ∂/∂x (e^{x+y}) + ∂/∂x (ln(xy)) = e^{x+y} + 1/x. Similarly v_y = e^{x+y} + 1/y. Hence v_x + v_y = 2 e^{x+y} + 1/x + 1/y.

w = x^x y^y = e^{x ln x + y ln y}. ∂w/∂x = w · ∂/∂x(x ln x + y ln y) = w (ln x + 1). Thus ∂w/∂x = x^x y^y(ln x + 1).

dS = S'(x_0) dx = 12 x_0 dx, so the approximate change is 12 x_0 dx.

dg/dt = g_x x' + g_y y'. Here g_x = 3x^2+10x, g_y = -4y+3y^2, x'=e^t, y'=-sin t. Substitute x=e^t,y=cos t to get dg/dt = (3e^{2t}+10e^t)e^t +(-4 cos t+3 cos^2 t)(-sin t) = 3e^{3t}+10e^{2t}+4 cos t sin t -3 cos^2 t sin t.